Circuit question -- Two sources and two resistors....

[Joshy]

Let me also point out that the main reason you are getting stuck on this problem is because you are taking Vo as the voltage at that top node which it is not. The left going current is neither Vo/4 nor Vo/(4 + 6). The correct (left going) current is simply V/(4 + 6) where V, as above, is simply the voltage at that essential node on top.

This thread is super old. Someone has already answered the question and another person bumped it way later probably not reading the old responses.

I wouldn't have followed up, but I did want to add something: I understand that you are correct about the "left going" current not being Vo/4, but the current in that left branch IS Vo/4 and can still be easily used for KCL. If it's really important you can say the "left going" current is -Vo/4. Sure you don't know what the voltage is at the top, but you can use Ohm's law on that resistor that has Vo assigned across it. This works because the two resistors are in series, and so you know the same current is flowing through it. So long as current is flowing through that branch the 6 Ohm resistor does not affect Vo.

The eye opener to that problem is that it would affect the power (second part of the problem). Finding the voltage at the top node would then become helpful because the second part of the question asks to find the power. I think considering Vo/4 from the start is easier and you can use that current to easily get the voltages without considering a voltage divider or careful bookkeeping of the polarities.

Spoiler

 

[boo]

Whoa! Didn't realize that this was an ancient thread. Would not waded in. But agree that left going current is - Vo/4.